If two tangents can be drawn to distinct branches of hyperbola $x^2-\frac{y^2}{4}=1$ from the point $(a,a^2)$, then pick the options

Question

$ a\in (-2,0)$

$a\in (0,2)$

$a\in (-\infty, -2)$

$a\in (\infty, 2)$

The equation for tangent is

$$a^2=ma \pm \sqrt{m^2-4}$$

$$m^2-4=a^4+m^2a^2-2ma^3$$

$$m^2(1-a^2)+2ma^3-(4+a^4)=0$$

For two distinct values

$$a^6+(1-a^2)(a^4+4)>0$$

$$a^4-4a^2+4>0$$ $$(a^2-2)^2>0$$

So $a>\sqrt 2$ and $a<-\sqrt 2$

But this doesn’t match with any of the given options. What’s am I doing wrong?

Answer 1

You’ve drawn the wrong conclusion from the last inequality. The square of any real number is nonnegative, so for any $a\ne\pm\sqrt2$, $(a^2-2)^2\gt0$.

The more fundamental problem with your approach to a solution, though, is that you’re looking at the wrong condition. You’re looking for points on the parabola $y=x^2$ though which two tangents can be drawn, but two tangents can be drawn to a hyperbola through any point that lies between its branches except along its asymptotes (and even there, if you consider the asymptotes to be tangents at infinity). The only two points on the parabola that aren’t in between the hyperbola’s branches are its two points of intersection with it, corresponding to $a=\pm\sqrt2$. What you really need to consider instead is where the points of tangency lie. Since this hyperbola’s branches lie on either side of the $y$-axis, this means that the $x$-coordinates of the points of contact should have opposite signs.

The extension of the chord of contact is the polar of $(a,a^2)$, namely $$ax-{a^2y\over4}=1.$$ Find the intersections of this line with the hyperbola and then find the conditions on $a$ for the product (or ratio, whichever gives you a simpler expression to work with) of their $x$-coordinates to be negative. You’ll need to take some care with values of $a$ for which either expression vanishes since by examining the product or ratio you’re effectively using the reciprocal of one or the other of them.

Answer 2

Note that the asymptotes of $x^2-\frac{y^2}{4}=1$ are $y=\pm 2x$, which intersect $y=x^2$, the curve of $(a,a^2)$ at $x=\pm2$.

Geometrically, for any point $(a,a^2)$ above the asymptotes, there is only one tangent line that can be drawn to either branch of the hyperbola, which can also be seen from that $a^2=ma \pm \sqrt{m^2-4}$ has only one root $m$ for each branch for $|a| >2$.

Thus, the options are $a\in(-\infty,-2)$ and $a\in (2,\infty)$.